What Is a Wilson Score Interval?
When you observe x successes in n independent binary trials, you know p̂ = x/n — the sample proportion. What you want to estimate is p, the underlying population proportion. A confidence interval gives you a plausible range for p based on your sample.
The most familiar formula, the Wald interval, is simply p̂ ± z × √(p̂(1−p̂)/n). It is intuitive and quick to compute, but it has a well-documented problem: when p̂ is near 0 or 1, or when n is small, the normal approximation it relies on is poor. The interval can produce impossible bounds — a negative lower bound or an upper bound above 1 — and its actual coverage can fall well below the stated confidence level.
Edwin Bidwell Wilson introduced a better approach in 1927. Instead of centering the interval at p̂, the Wilson method solves for which values of p would not be rejected by a score test at level α. The result is an interval that stays within [0, 1] automatically and has coverage properties that are generally closer to the nominal level, particularly in the small-sample and boundary-proportion situations where the Wald method struggles most.
The Wilson interval is the default method used by many statistical packages, including R's prop.test(), and is recommended in several applied statistics texts. Coverage properties across different methods are discussed in detail in Agresti and Coull (1998) and Brown, Cai, and DasGupta (2001) in the Statistical Science journal. The underlying probability framework draws on the binomial distribution.
The Wilson score interval is a confidence interval for a proportion that handles small samples and extreme proportions better than the Wald interval. Use it whenever you are estimating a proportion from count data and want bounds that stay within [0, 1].
Wilson Score Interval Formula
The two bounds of the Wilson interval share a common structure. Let x be the number of successes, n the total number of trials, p̂ = x/n the observed proportion, and z the standard normal critical value for the chosen confidence level.
Upper = [ p̂ + z²/(2n) + z·√(p̂(1−p̂)/n + z²/(4n²)) ] / [ 1 + z²/n ]
x = number of successes
n = total trials
p̂ = x / n
z = critical value
z² = z × z
You can also think of the formula in two parts: a center and a margin.
- Center = (p̂ + z²/(2n)) / (1 + z²/n) — this is a weighted average of p̂ and 0.5, pulled slightly toward 0.5 compared to the raw observed proportion
- Margin = z × √(p̂(1−p̂)/n + z²/(4n²)) / (1 + z²/n) — the half-width of the interval around the center
- Lower = Center − Margin
- Upper = Center + Margin
What Do the Variables in the Formula Mean?
| Variable | Meaning | Example (60/100) |
|---|---|---|
| x | Number of observed successes (count, not percentage) | 60 |
| n | Total number of observations (trials) | 100 |
| p̂ | Observed sample proportion = x/n | 0.60 |
| 1 − p̂ | Observed proportion of failures | 0.40 |
| z | Standard normal critical value for the chosen confidence level | 1.960 (for 95%) |
| z² | Square of the critical value | 3.8416 |
| z²/(2n) | Continuity shift applied to p̂ in the numerator | 0.01921 |
| z²/n | Denominator adjustment that shrinks the interval | 0.03842 |
| 1 + z²/n | The denominator; always greater than 1 | 1.03842 |
Critical Values for Common Confidence Levels
The z critical value comes from the standard normal distribution. For a two-sided confidence interval at level C, z is the value such that P(−z ≤ Z ≤ z) = C for a standard normal random variable Z. The values most often used in practice are:
| Confidence Level | z Critical Value | z² | z²/(2n) for n = 100 |
|---|---|---|---|
| 90% | 1.645 | 2.7060 | 0.01353 |
| 95% | 1.960 | 3.8416 | 0.01921 |
| 99% | 2.576 | 6.6355 | 0.03318 |
These are symmetric two-sided critical values. A 95% confidence interval uses z = 1.960; a 99% interval uses z = 2.576. You can read these from a z-table or compute them from the inverse standard normal function.
How to Calculate a Wilson Score Interval by Hand
The process below applies to every worked example that follows. Keep enough decimal places during intermediate steps — rounding too early introduces errors that compound. Round only the final bounds to a sensible number of decimal places.
Count x (successes) and n (total observations)
Make sure x is a raw count, not a percentage. n is the total number of trials, not the number of failures.
Calculate p̂ = x / n
This is the observed proportion. If you have percentages, convert them to decimals first (e.g., 60% → 0.60).
Choose the confidence level and find z
95% → z = 1.960. 90% → z = 1.645. 99% → z = 2.576. Compute z² and store it.
Calculate the denominator: 1 + z²/n
Divide z² by n, then add 1. This term appears in both bounds.
Calculate the center numerator: p̂ + z²/(2n)
Divide z² by 2n (not n), then add p̂. This is the numerator before adjusting for the margin.
Calculate the term inside the square root: p̂(1−p̂)/n + z²/(4n²)
Compute each part separately. p̂(1−p̂)/n is the variance contribution; z²/(4n²) is the continuity term. Add them together before taking the square root.
Multiply z by the square root, then divide by the denominator
This gives the margin M. Lower = (center numerator − z × √(…)) / denominator. Upper = (center numerator + z × √(…)) / denominator. Equivalently: Lower = Center − M, Upper = Center + M where Center = center numerator / denominator.
Verify: 0 ≤ Lower ≤ Upper ≤ 1
A valid Wilson interval always satisfies this. If either bound falls outside [0, 1], check your arithmetic — the Wilson formula guarantees valid bounds when computed correctly.
Interpret the result
State the confidence level and the bounds. Avoid saying "there is a 95% probability the true proportion is in this range" — see the interpretation section below for the correct frequentist phrasing.
The interactive calculator in the sidebar lets you compute a Wilson interval quickly. It is worth doing the arithmetic by hand at least once so the steps are clear when you need to explain or audit results.
Worked Example 1: 60 Successes out of 100
A production line records 60 conforming parts out of 100 inspected. Calculate the 95% Wilson confidence interval for the true conformance rate.
60 successes from 100 trials — 95% confidence interval
Identify: x = 60, n = 100
Calculate p̂: p̂ = 60/100 = 0.60 | 1 − p̂ = 0.40
Select z: 95% confidence → z = 1.960, so z² = 3.8416
Denominator: 1 + z²/n = 1 + 3.8416/100 = 1 + 0.038416 = 1.038416
Center numerator: p̂ + z²/(2n) = 0.60 + 3.8416/200 = 0.60 + 0.019208 = 0.619208
Inside the root: p̂(1−p̂)/n = (0.60 × 0.40)/100 = 0.0024 | z²/(4n²) = 3.8416/40000 = 0.000096 | Sum = 0.002496
Square root: √0.002496 = 0.049960
z × √(…): 1.960 × 0.049960 = 0.097922
Lower numerator: 0.619208 − 0.097922 = 0.521286 | Lower bound: 0.521286 / 1.038416 = 0.5020
Upper numerator: 0.619208 + 0.097922 = 0.717130 | Upper bound: 0.717130 / 1.038416 = 0.6906
✅ 95% Wilson CI: (0.502, 0.691) — equivalently (50.2%, 69.1%). The data are consistent with a true conformance rate between roughly 50% and 69%.
Worked Example 2: Small Sample (7 out of 10)
A quality engineer tests 10 widgets and 7 pass. What does the 95% Wilson interval look like, and how does it compare to the Wald result?
7 successes from 10 trials — 95% confidence interval
Identify: x = 7, n = 10, p̂ = 0.70, 1 − p̂ = 0.30
z = 1.960, z² = 3.8416
Denominator: 1 + 3.8416/10 = 1.38416
Center numerator: 0.70 + 3.8416/20 = 0.70 + 0.19208 = 0.89208
Inside root: (0.70 × 0.30)/10 + 3.8416/400 = 0.021 + 0.009604 = 0.030604
z × √(0.030604): 1.960 × 0.17494 = 0.34288
Lower: (0.89208 − 0.34288) / 1.38416 = 0.54920 / 1.38416 = 0.3967
Upper: (0.89208 + 0.34288) / 1.38416 = 1.23496 / 1.38416 = 0.8923
✅ Wilson 95% CI: (0.397, 0.892). For comparison, the Wald interval is 0.70 ± 1.960 × 0.1449 = (0.416, 0.984). The Wald upper bound (0.984) is plausible, but the Wald misses the asymmetry that is natural when p̂ = 0.70 with only 10 observations.
With n = 10, the normal approximation underlying the Wald formula is not reliable. The Wilson interval is wider and shifted, reflecting genuine uncertainty rather than an artificially symmetric range. This is the case where the Wilson approach earns its keep.
Worked Example 3: Low Proportion (2 out of 50)
A lab records 2 defective units out of 50 tested. Calculate the 95% Wilson confidence interval for the defect rate.
2 successes from 50 trials — 95% confidence interval
Identify: x = 2, n = 50, p̂ = 2/50 = 0.04, 1 − p̂ = 0.96
z = 1.960, z² = 3.8416
Denominator: 1 + 3.8416/50 = 1 + 0.076832 = 1.076832
Center numerator: 0.04 + 3.8416/100 = 0.04 + 0.038416 = 0.078416
Inside root: (0.04 × 0.96)/50 + 3.8416/10000 = 0.000768 + 0.00038416 = 0.00115216
z × √(0.00115216): 1.960 × 0.033944 = 0.066530
Lower: (0.078416 − 0.066530) / 1.076832 = 0.011886 / 1.076832 = 0.0110
Upper: (0.078416 + 0.066530) / 1.076832 = 0.144946 / 1.076832 = 0.1346
✅ Wilson 95% CI: (0.011, 0.135) — that is, (1.1%, 13.5%). The interval is asymmetric: it extends 2.9 percentage points below p̂ = 4% but 9.5 points above it. The Wald interval would be 0.04 ± 0.055 = (−0.015, 0.095), giving an impossible negative lower bound.
Worked Example 4: High Proportion (48 out of 50)
A component passes inspection 48 times out of 50. Calculate the 95% Wilson interval for the true pass rate.
48 successes from 50 trials — 95% confidence interval
Identify: x = 48, n = 50, p̂ = 0.96, 1 − p̂ = 0.04
z = 1.960, z² = 3.8416
Denominator: 1.076832 (same as Example 3)
Center numerator: 0.96 + 0.038416 = 0.998416
Inside root: (0.96 × 0.04)/50 + 3.8416/10000 = 0.000768 + 0.00038416 = 0.00115216
z × √(0.00115216): 1.960 × 0.033944 = 0.066530
Lower: (0.998416 − 0.066530) / 1.076832 = 0.931886 / 1.076832 = 0.8654
Upper: (0.998416 + 0.066530) / 1.076832 = 1.064946 / 1.076832 = 0.9890
✅ Wilson 95% CI: (0.865, 0.989). Compare with the Wald interval: 0.96 ± 0.055 = (0.905, 1.015) — the Wald upper bound exceeds 1, which is impossible. The Wilson interval remains valid.
Worked Example 5: Zero Successes (0 out of 20)
A rare event is observed 0 times in 20 trials. This is the edge case where the Wald interval collapses to [0, 0]. The Wilson interval still gives a meaningful result.
0 successes from 20 trials — 95% confidence interval
Identify: x = 0, n = 20, p̂ = 0, 1 − p̂ = 1
z = 1.960, z² = 3.8416
Denominator: 1 + 3.8416/20 = 1 + 0.19208 = 1.19208
Center numerator: 0 + 3.8416/40 = 0 + 0.09604 = 0.09604
Inside root: (0 × 1)/20 + 3.8416/1600 = 0 + 0.002401 = 0.002401
z × √(0.002401): 1.960 × 0.04900 = 0.09604
Lower: (0.09604 − 0.09604) / 1.19208 = 0 / 1.19208 = 0.0000
Upper: (0.09604 + 0.09604) / 1.19208 = 0.19208 / 1.19208 = 0.1612
✅ Wilson 95% CI: (0.000, 0.161). With 0 successes in 20 trials, we cannot rule out a true rate as high as 16.1%. The Wald interval would be [0, 0], which incorrectly suggests the true rate is certainly zero.
When p̂ = 0, the Wilson lower bound is exactly 0. This does not mean the true rate is zero — it means the data are consistent with zero, but the upper bound shows the rate could plausibly be as high as 16.1%. The interval [0, 0.161] is far more informative than [0, 0].
Worked Example 6: All Successes (20 out of 20)
20 successes from 20 trials — 95% confidence interval
Identify: x = 20, n = 20, p̂ = 1.0, 1 − p̂ = 0
Denominator: 1.19208 (same as Example 5)
Center numerator: 1.0 + 0.09604 = 1.09604
Inside root: (1.0 × 0)/20 + 0.002401 = 0 + 0.002401 = 0.002401
z × √(0.002401): 1.960 × 0.04900 = 0.09604
Lower: (1.09604 − 0.09604) / 1.19208 = 1.00000 / 1.19208 = 0.8388
Upper: (1.09604 + 0.09604) / 1.19208 = 1.19208 / 1.19208 = 1.0000
✅ Wilson 95% CI: (0.839, 1.000). Observing 20 successes from 20 trials means the upper bound hits 1.0 exactly. The lower bound (0.839) reflects genuine sampling uncertainty — a true rate below 83.9% would have been unlikely to produce 20-out-of-20.
Worked Example 7: 95% vs 99% Confidence
Using the same data (60 successes from 100 trials), compare how the interval changes when confidence changes from 95% to 99%.
60/100 — comparing 95% and 99% intervals
95% (z = 1.960, already computed in Example 1): Lower = 0.502, Upper = 0.691
99% setup: z = 2.576, z² = 6.6355
Denominator (99%): 1 + 6.6355/100 = 1.066355
Center numerator (99%): 0.60 + 6.6355/200 = 0.60 + 0.033178 = 0.633178
Inside root (99%): 0.0024 + 6.6355/40000 = 0.0024 + 0.000166 = 0.002566
z × √(0.002566) (99%): 2.576 × 0.050655 = 0.130487
Lower (99%): (0.633178 − 0.130487) / 1.066355 = 0.502691 / 1.066355 = 0.4714
Upper (99%): (0.633178 + 0.130487) / 1.066355 = 0.763665 / 1.066355 = 0.7161
✅ 95% CI: (0.502, 0.691) — width = 0.189. | 99% CI: (0.471, 0.716) — width = 0.245. Higher confidence requires a wider interval. You gain certainty at the cost of precision.
Worked Example 8: Different Sample Sizes
The same proportion (60%) observed with three different sample sizes shows how increasing n narrows the interval.
| Successes | Trials (n) | p̂ | Wilson Lower | Wilson Upper | Width |
|---|---|---|---|---|---|
| 30 | 50 | 0.60 | 0.463 | 0.724 | 0.261 |
| 300 | 500 | 0.60 | 0.558 | 0.641 | 0.083 |
| 3,000 | 5,000 | 0.60 | 0.586 | 0.614 | 0.028 |
The calculation for 30/50 follows the same steps as Example 1. Intermediate values: denominator = 1 + 3.8416/50 = 1.076832; center num = 0.60 + 0.038416 = 0.638416; inside root = (0.24)/50 + 3.8416/10000 = 0.0048 + 0.00038416 = 0.00518416; z × √(…) = 1.960 × 0.071982 = 0.141045; lower = (0.638416 − 0.141045)/1.076832 = 0.4628; upper = (0.638416 + 0.141045)/1.076832 = 0.7237. The pattern is clear: as n increases ten-fold, the interval width roughly decreases by a factor of √10.
Worked Example 9: Business Conversion Rate
A website records 125 conversions from 500 visitors during a campaign. Calculate the 95% Wilson interval for the true conversion rate. (Data are hypothetical.)
125 conversions from 500 visitors — 95% confidence interval
Identify: x = 125, n = 500, p̂ = 125/500 = 0.25, 1 − p̂ = 0.75
z = 1.960, z² = 3.8416
Denominator: 1 + 3.8416/500 = 1.007683
Center numerator: 0.25 + 3.8416/1000 = 0.25 + 0.003842 = 0.253842
Inside root: (0.25 × 0.75)/500 + 3.8416/1000000 = 0.000375 + 0.0000038 = 0.0003788
z × √(0.0003788): 1.960 × 0.019463 = 0.038148
Lower: (0.253842 − 0.038148) / 1.007683 = 0.215694 / 1.007683 = 0.2141
Upper: (0.253842 + 0.038148) / 1.007683 = 0.291990 / 1.007683 = 0.2898
✅ Wilson 95% CI: (0.214, 0.290) — that is, (21.4%, 29.0%). A reasonable planning assumption for the true conversion rate lies in this range. At n = 500, the interval is fairly narrow, so this estimate is reasonably precise.
Worked Example 10: A/B Testing
Two website variants are tested. Version A has 45 conversions from 500 visitors; Version B has 60 conversions from 500 visitors. Compute the Wilson 95% interval for each separately.
Version A: 45/500 | Version B: 60/500
Version A: p̂ = 45/500 = 0.09. Following the same steps: denominator = 1.007683; center num = 0.09 + 0.003842 = 0.093842; inside root = (0.09 × 0.91)/500 + 0.0000038 = 0.0001638 + 0.0000038 = 0.0001676; z × √(0.0001676) = 1.960 × 0.012947 = 0.025376; Lower = (0.093842 − 0.025376)/1.007683 = 0.0679; Upper = (0.093842 + 0.025376)/1.007683 = 0.1183
Version B: p̂ = 60/500 = 0.12. Inside root = (0.12 × 0.88)/500 + 0.0000038 = 0.0002112 + 0.0000038 = 0.000215; z × √(0.000215) = 1.960 × 0.014663 = 0.028739; Lower = (0.12 + 0.003842 − 0.028739)/1.007683 = 0.095103/1.007683 = 0.0944; Upper = (0.123842 + 0.028739)/1.007683 = 0.152581/1.007683 = 0.1514
Version A: (6.8%, 11.8%) | Version B: (9.4%, 15.1%). The intervals overlap, but overlapping confidence intervals are not a valid substitute for a formal comparison test. To determine whether Version B is significantly better, use a two-proportion z-test or chi-square test. See the hypothesis testing guide for that approach.
Even when two Wilson intervals overlap, the difference between the two proportions may still be statistically significant. Use a two-proportion z-test or chi-square test for the formal comparison — the individual Wilson intervals estimate each proportion separately, not their difference.
Worked Example 11: Customer Satisfaction
A company surveys 100 customers and 84 report satisfaction. Estimate the 95% Wilson interval for the true satisfaction rate.
84 satisfied from 100 respondents — 95% confidence interval
Identify: x = 84, n = 100, p̂ = 0.84, 1 − p̂ = 0.16
Denominator: 1.038416
Center numerator: 0.84 + 0.019208 = 0.859208
Inside root: (0.84 × 0.16)/100 + 0.000096 = 0.001344 + 0.000096 = 0.001440
z × √(0.001440): 1.960 × 0.037947 = 0.074375
Lower: (0.859208 − 0.074375) / 1.038416 = 0.784833 / 1.038416 = 0.7558
Upper: (0.859208 + 0.074375) / 1.038416 = 0.933583 / 1.038416 = 0.8990
✅ Wilson 95% CI: (0.756, 0.899). While 84% of respondents expressed satisfaction, the uncertainty in this survey means the true rate could be as low as 75.6% or as high as 89.9%. This interval is more useful for decision-making than the point estimate alone.
Worked Example 12: Defect Rate
A production run of 200 units contains 8 defectives. Estimate the 95% Wilson interval for the true defect rate.
8 defectives from 200 units — 95% confidence interval
Identify: x = 8, n = 200, p̂ = 8/200 = 0.04, 1 − p̂ = 0.96
Denominator: 1 + 3.8416/200 = 1.019208
Center numerator: 0.04 + 3.8416/400 = 0.04 + 0.009604 = 0.049604
Inside root: (0.04 × 0.96)/200 + 3.8416/160000 = 0.000192 + 0.000024 = 0.000216
z × √(0.000216): 1.960 × 0.014697 = 0.028807
Lower: (0.049604 − 0.028807) / 1.019208 = 0.020797 / 1.019208 = 0.0204
Upper: (0.049604 + 0.028807) / 1.019208 = 0.078411 / 1.019208 = 0.0769
✅ Wilson 95% CI: (0.020, 0.077). The defect rate estimate is 4%, but the true rate could plausibly range from 2.0% to 7.7%. This interval informs how tight a quality specification needs to be before the current process is confidently within tolerance.
Worked Example 13: Wilson Score Ranking
The Wilson lower bound is sometimes used in ranking systems for items with positive/negative ratings. The idea: instead of ranking by raw percentage, rank by the lower bound of the Wilson interval. This prevents items with very few ratings but a high percentage from ranking above items with many consistent ratings.
Consider two products. Product A has 90 positive ratings from 100 total (90%). Product B has 9 positive ratings from 10 total (90%). The raw percentage is identical, but Product B's estimate is far less certain.
Comparing two items with the same percentage but different sample sizes
Product A: x = 90, n = 100, p̂ = 0.90. Denominator = 1.038416; center num = 0.90 + 0.019208 = 0.919208; inside root = (0.90 × 0.10)/100 + 0.000096 = 0.000900 + 0.000096 = 0.000996; z × √(0.000996) = 1.960 × 0.031559 = 0.061855; Lower A = (0.919208 − 0.061855) / 1.038416 = 0.857353 / 1.038416 = 0.8257
Product B: x = 9, n = 10, p̂ = 0.90. From Example 2 structure: denominator = 1.38416; center num = 0.90 + 0.19208 = 1.09208; inside root = (0.90 × 0.10)/10 + 3.8416/400 = 0.009 + 0.009604 = 0.018604; z × √(0.018604) = 1.960 × 0.136396 = 0.267336; Lower B = (1.09208 − 0.267336) / 1.38416 = 0.824744 / 1.38416 = 0.5959
✅ Product A lower bound: 0.826. Product B lower bound: 0.596. Although both have 90% positive ratings, ranking by Wilson lower bound places Product A clearly above Product B, reflecting its much greater number of reviews.
The Wilson score interval is a confidence interval for an estimated proportion. Using the lower bound to rank items is a separate application — it exploits the property that the lower bound is conservative when sample size is small. The two uses are related but distinct. The ranking application does not give you a formal probability statement about ordering.
Wilson vs Wald: Numerical Comparison
The Wald interval for a proportion is simply p̂ ± z × √(p̂(1−p̂)/n). It is quick to compute but has two known problems: it can produce bounds outside [0, 1], and its actual coverage probability is often below the stated confidence level when n is small or p is extreme.
| Scenario | p̂ | n | Wilson Lower | Wilson Upper | Wald Lower | Wald Upper |
|---|---|---|---|---|---|---|
| Example 1 | 0.60 | 100 | 0.502 | 0.691 | 0.504 | 0.696 |
| Small sample | 0.70 | 10 | 0.397 | 0.892 | 0.416 | 0.984 |
| Low proportion | 0.04 | 50 | 0.011 | 0.135 | −0.015 | 0.095 |
| High proportion | 0.96 | 50 | 0.865 | 0.989 | 0.905 | 1.015 |
| Zero successes | 0.00 | 20 | 0.000 | 0.161 | 0.000 | 0.000 |
| All successes | 1.00 | 20 | 0.839 | 1.000 | 1.000 | 1.000 |
For large n and p̂ away from 0 and 1 (like Example 1), Wilson and Wald produce very similar results. The differences become meaningful when n is small, or when p̂ is close to 0 or 1. The Wald produces an impossible lower bound of −1.5% for the low-proportion case, and collapses entirely for zero successes. Wilson handles both correctly.
Wilson vs Other Methods
| Method | Approach | Small-sample behavior | Boundary behavior | Typical use |
|---|---|---|---|---|
| Wald | Normal approximation centered at p̂ | Poor; can give impossible bounds | Fails at p̂ = 0 or 1 | Textbook introduction only |
| Wilson | Score-test inversion | Good; bounds always in [0,1] | Handles extremes correctly | General use, default in R |
| Agresti-Coull | Adjusted p̃ = (x+2)/(n+4), then Wald | Good approximation to Wilson | Good | Simple hand calculation |
| Clopper-Pearson | Exact binomial; inverts cumulative probabilities | Conservative (wider than needed) | Exact at boundaries | Regulatory/safety contexts requiring exact bounds |
| Jeffreys | Bayesian with Jeffreys prior | Good; similar to Wilson | Handles boundaries well | Bayesian analysis |
No single method is best for every situation. Wilson is a reasonable general choice. For regulatory contexts where conservative bounds are required, Clopper-Pearson is often preferred. When hand calculation is needed without iterative computing, the Agresti-Coull approximation is easy to apply. The choice of method should depend on the application, not just convenience. Coverage properties of these methods are compared in detail in the Brown, Cai, and DasGupta (2001) paper in Statistical Science.
How to Interpret a Wilson Score Interval
Confidence intervals for proportions are frequently misinterpreted. The correct frequentist interpretation is:
This is a statement about the long-run performance of the procedure, not a probability statement about where p lies after you have observed your data. Once you have computed a specific interval, the true proportion is either inside it or outside it — there is no probability involved for that particular interval.
Common incorrect phrasings to avoid:
- "There is a 95% probability that p is between 0.502 and 0.691." — incorrect; p is a fixed unknown value
- "We are 95% sure the true rate is in this range." — colloquial and misleading
Acceptable phrasings:
- "The 95% Wilson confidence interval for p is (0.502, 0.691)."
- "Using the Wilson method, a 95% confidence interval for the true proportion is (0.502, 0.691)."
- "The data are consistent with a true rate between 50.2% and 69.1% at the 95% confidence level."
Why Is the Wilson Interval Often Asymmetric?
The Wald interval is symmetric by construction — it extends the same distance below and above p̂. The Wilson interval is not, and this asymmetry is a deliberate consequence of how it is derived.
When p̂ is close to 0, the distribution of the count X under the binomial model is right-skewed. A symmetric interval would extend into negative territory, which is impossible. The Wilson method effectively corrects for this by shifting the center slightly away from the boundary and shrinking the lower margin. The same applies in mirror image when p̂ is close to 1.
Even at moderate p̂ values with small n, the normal approximation on which symmetric intervals rely is imperfect, and the Wilson interval captures some of this non-normality through its different center. As n increases and p̂ moves away from 0 and 1, the Wilson interval becomes more symmetric — the two methods converge. The asymmetry is most visible, and most important, precisely where the Wald interval is least reliable.
Assumptions and Limitations
- Binary outcomes: Each observation must have exactly two possible results (success or failure). The Wilson interval is not appropriate for count data, continuous measurements, or ordered categories.
- Independence: Observations must be independent. If responses are correlated — for example, responses from the same household, or repeated measurements from the same unit — the binomial model does not apply without adjustment.
- Random sampling: The sample should be representative of the population of interest. A mathematically correct Wilson interval does not compensate for biased sampling.
- Fixed n: The interval assumes n is fixed in advance, not determined by the data (as in sequential testing). If n is data-dependent, the coverage properties change.
- No clustering: If data are clustered (e.g., students within classrooms), a simple binomial interval understates uncertainty. Use survey-weighted methods instead.
A key limitation worth emphasizing: the Wilson interval quantifies sampling uncertainty only. It does not account for measurement error, response bias, selection bias, or any other non-sampling error. An interval can be mathematically precise while still being misleading if the data were collected poorly. The study design matters as much as the analysis method.
Common Mistakes
| Mistake | What goes wrong | How to fix it |
|---|---|---|
| Entering a percentage as p̂ | Using p̂ = 60 instead of 0.60 | Always use the decimal form; 60% = 0.60 |
| Confusing x with n | Swapping successes and total trials | x is the count of successes; n is the total including failures |
| Using the wrong z value | Using z = 1.645 (90%) for a 95% interval | 95% → z = 1.960; 99% → z = 2.576 |
| Rounding intermediate steps | Rounding to 2 decimal places mid-calculation | Keep at least 6 significant figures until the final bounds |
| Interpreting a narrow interval as unbiased | Assuming a small interval means accurate measurement | A narrow interval means precise sampling only; bias is a separate issue |
| Using overlapping intervals as a significance test | Concluding no significant difference because two intervals overlap | Use a two-proportion test; overlapping intervals are not equivalent to non-significance |
| Confusing Wilson interval with Wilson ranking | Treating the lower bound used for ranking as a probability statement | The lower bound ranking is a separate heuristic, not a formal interval |
| Ignoring dependence | Applying Wilson to clustered or repeated data | Use methods that account for the correlation structure |
Wilson Score Interval Calculator
Use this calculator to verify your worked examples or check new data. The results match the formula above exactly.
Wilson Score Interval Calculator
Practical Applications
Marketing and A/B Testing
Estimating click-through rates, conversion rates, or open rates from campaign data.
Quality Control
Bounding defect rates and pass rates from inspection samples in manufacturing.
Survey Research
Expressing uncertainty around agreement rates, satisfaction scores, or yes/no responses.
Ratings and Reviews
Ranking products or content using the Wilson lower bound to account for sample size differences.
Clinical and Public Health
Estimating response rates, prevalence, and diagnostic accuracy from study data.
Machine Learning
Bounding classification accuracy, precision, and recall estimated from held-out test sets.
Frequently Asked Questions
A confidence interval for a population proportion derived by inverting the score test. It generally behaves better than the Wald interval for small samples and proportions near 0 or 1 because it always produces bounds within [0, 1].
The Wald interval can produce impossible bounds below 0 or above 1 and has poor coverage when n is small or p̂ is near 0 or 1. The Wilson interval stays within [0, 1] and has better actual coverage in these situations. For large n and p̂ away from the boundaries, both methods give similar results.
For a two-sided 95% interval, use z = 1.960. For 90% use z = 1.645. For 99% use z = 2.576. These values come from the standard normal distribution.
The Wilson formula still produces a meaningful interval. The lower bound is 0 because p̂ = 0, but the upper bound is positive and reflects the maximum true rate that would have been plausible given no observed successes. The Wald interval, by contrast, collapses to [0, 0], which is misleading.
Because the Wilson interval does not simply extend the same distance above and below p̂. The center of the interval is shifted slightly toward 0.5, and the margin is not symmetric when p̂ is close to 0 or 1. This reflects the fact that proportions are bounded between 0 and 1. Symmetry is more appropriate near p̂ = 0.5 with large n.
No. A narrow interval means the sampling variability is small, typically because n is large. It says nothing about whether the data were collected without bias, whether the sample was representative, or whether the definition of success was consistent. Precision and accuracy are different things.
No. Overlapping individual confidence intervals do not imply a non-significant difference between proportions. For a formal comparison, use a two-proportion z-test or a chi-square test. The individual intervals estimate each proportion separately and are not designed to test the difference directly.
Larger n produces a narrower interval. For a fixed observed proportion, the interval width approximately halves each time n quadruples. For example, intervals for the same p̂ = 0.60 become progressively narrower as the sample size increases from n = 50 to n = 500 and n = 5000.
Not exactly. The frequentist interpretation is that the interval-construction procedure will contain the true proportion in 95% of repeated samples from the same population. Once a specific interval is computed, the true proportion is either inside it or not. Probability does not apply to that fixed interval in the frequentist interpretation.
Ranking items such as products, reviews, or posts by the Wilson lower bound rather than their raw percentage gives a conservative score that accounts for sample size. An item with 90% positive ratings from 10 reviews ranks below one with 90% from 100 reviews because the first estimate is much less certain.
Agresti-Coull adds 2 successes and 2 failures to the data, giving p̃ = (x + 2)/(n + 4), and then applies the Wald formula to p̃. The result closely approximates the Wilson interval and is slightly simpler to compute by hand. They are not identical, but they perform similarly in many situations.
No. The Wilson interval is specifically designed for binary, two-outcome data. For multiple categories, use multinomial confidence intervals or apply the Wilson formula separately to each category proportion by treating each category as a binary question, with appropriate adjustment for multiple comparisons when needed.
The difference is most meaningful when n is small, roughly below 30; when p̂ is below 0.10 or above 0.90; or when x = 0 or x = n. In these situations, the Wald interval can produce poor or impossible bounds, making the Wilson interval clearly preferable.
Key Takeaways
- Formula: Lower/Upper = [p̂ + z²/(2n) ± z√(p̂(1−p̂)/n + z²/(4n²))] / (1 + z²/n)
- z values: 1.645 (90%), 1.960 (95%), 2.576 (99%)
- Key property: Bounds always lie in [0, 1] — unlike the Wald interval
- Best when: n is small, p̂ is near 0 or 1, or data include zero/all successes
- Interpretation: Frequentist — 95% of such intervals contain the true p in repeated sampling
- Not for: Testing differences between proportions; use a two-proportion test for that
- Limitations: Assumes independent binary observations from a representative sample
Further reading on interval methods for proportions: the NIST Engineering Statistics Handbook covers confidence intervals for proportions in detail. Coverage analysis for all major methods appears in Brown, Cai, and DasGupta (2001) in Statistical Science. For related methods, the confidence intervals section of Statistics Fundamentals covers confidence interval concepts more broadly.